examples of selecting a ball screw - thk · pdf filescrew shaft diameter of 30 mm and a lead...

Ball Screw

Examples of Selecting a Ball Screw High-speed Transfer Equipment (Horizontal Use)

[Selection Conditions] Table Mass m 1 =60kg Work Mass m 2 =20kg Stroke length ℓ S =1000mm Maximum speed V max =1m/s Acceleration time t 1 = 0.15s Deceleration time t 3 = 0.15s Number of reciprocations per minute n =8min -1 Backlash 0.15mm Positioning accuracy 0.3 mm/1000 mm (Perform positioning from the negative direction)

Positioning accuracy repeatability 0.1 mm Minimum feed amount s = 0.02mm/pulse Desired service life time 30000h Driving motor AC servo motor Rated rotational speed: 3,000 min -1 Inertial moment of the motor J m =1×10 –3 kg•m 2 Reduction gear None (direct coupling)A=1 Frictional coeffi cient of the guide surface =0.003 (rolling) Guide surface resistance f=15 N (without load)

Work mass Table mass

Ball screw nut Ball screw shaft Motor

+

+ m2

m1

[Selection Items] Screw shaft diameter Lead Nut model No. Accuracy Axial clearance Screw shaft support method Driving motor

Point of SelectionExamples of Selecting a Ball Screw

B15-69

511E

[Selecting Lead Angle Accuracy and Axial Clearance] Selecting Lead Angle Accuracy

To achieve positioning accuracy of 0.3 mm/1,000 mm:

1000 300 = ±0.3 ±0.09

The lead angle accuracy must be 0.09 mm/300 mm or higher. Therefore, select the following as the accuracy grade of the Ball Screw (see Table1 on B15-20 ). C7 (travel distance error: 0.05mm/300mm) Accuracy grade C7 is available for both the Rolled and the Precision Ball Screws. Assume that a Rolled Ball Screw is selected here because it is less costly.

Selecting Axial Clearance To satisfy the backlash of 0.15 mm, it is necessary to select a Ball Screw with an axial clearance of 0.15 mm or less. Therefore, a Rolled Ball Screw model with a screw shaft diameter of 32 mm or less that meets the axial clearance of 0.15 mm or less (see Table13 on B15-27 ) meets the requirements. Thus, a Rolled Ball Screw model with a screw shaft diameter of 32 mm or less and an accuracy grade of C7 is selected.

[Selecting a Screw Shaft] Assuming the Screw Shaft Length

Assume the overall nut length to be 100 mm and the screw shaft end length to be 100 mm. Therefore, the overall length is determined as follows based on the stroke length of 1,000 mm. 1000 + 200 = 1200 mm Thus, the screw shaft length is assumed to be 1,200 mm.

Selecting a Lead With the driving motor’s rated rotational speed being 3,000 min -1 and the maximum speed 1 m/s, the Ball Screw lead is obtained as follows:

1×1000×60 3000

= 20 mm

Therefore, it is necessary to select a type with a lead of 20 mm or longer. In addition, the Ball Screw and the motor can be mounted in direct coupling without using a reduc-tion gear. The minimum resolution per revolution of an AC servomotor is obtained based on the resolution of the encoder (1,000 p/rev; 1,500 p/rev) provided as a standard accessory for the AC servomotor, as indicated below. 1000 p/rev(without multiplication) 1500 p/rev(without multiplication) 2000 p/rev(doubled) 3000 p/rev(doubled) 4000 p/rev(quadrupled) 6000 p/rev(quadrupled)

B15-70

511E

Ball Screw

To meet the minimum feed amount of 0.02 mm/pulse, which is the selection requirement, the follow-ing should apply. Lead 20mm —— 1000 p/rev 30mm —— 1500 p/rev 40mm —— 2000 p/rev 60mm —— 3000 p/rev 80mm —— 4000 p/rev

Selecting a Screw Shaft Diameter Those Ball Screw models that meet the requirements defi ned in Section [Selecting Lead Angle Accu-racy and Axial Clearance] on B15-70 : a rolled Ball Screw with a screw shaft diameter of 32 mm or less; and the requirement defined in Section [Selecting a Screw Shaft] on B15-70 : a lead of 20, 30, 40, 60 or 80 mm (see Table20 on B15-35 ) are as follows. Shaft diameter Lead 15mm —— 20mm 15mm —— 30mm 20mm —— 20mm 20mm —— 40mm 30mm —— 60mm Since the screw shaft length has to be 1,200 mm as indicated in Section [Selecting a Screw Shaft] on B15-70 , the shaft diameter of 15 mm is insufficient. Therefore, the Ball Screw should have a screw shaft diameter of 20 mm or greater. Accordingly, there are three combinations of screw shaft diameters and leads that meet the require-ments: screw shaft diameter of 20 mm/lead of 20 mm; 20 mm/40 mm; and 30 mm/60 mm.

Selecting a Screw Shaft Support Method Since the assumed type has a long stroke length of 1,000 mm and operates at high speed of 1 m/s, select either the fi xed-supported or fi xed-fi xed confi guration for the screw shaft support. However, the fi xed-fi xed confi guration requires a complicated structure, needs high accuracy in the installation. Accordingly, the fi xed-supported confi guration is selected as the screw shaft support method.


B15-71

511E

Studying the Permissible Axial Load Calculating the Maximum Axial Load

Guide surface resistance f=15 N (without load) Table Mass m 1 =60 kg Work Mass m 2 =20 kg Frictional coeffi cient of the guide surface = 0.003 Maximum speed V max =1 m/s Gravitational acceleration g = 9.807 m/s 2 Acceleration time t 1 = 0.15s

Accordingly, the required values are obtained as follows. Acceleration:

α = = 6.67 m/s2Vmax

t1

During forward acceleration: Fa 1 = • (m 1 + m 2 ) g + f + (m 1 + m 2 ) • = 550 N During forward uniform motion: Fa 2 = • (m 1 + m 2 ) g + f = 17 N During forward deceleration: Fa 3 = • (m 1 + m 2 ) g + f – (m 1 + m 2 ) • = –516 N During backward acceleration: Fa 4 = –• (m 1 + m 2 ) g – f – (m 1 + m 2 ) • = –550 N During uniform backward motion: Fa 5 = –• (m 1 + m 2 ) g – f = – 17 N During backward deceleration: Fa 6 = –• (m 1 + m 2 ) g – f + (m 1 + m 2 ) • = 516 N Thus, the maximum axial load applied on the Ball Screw is as follows: Fa max = Fa 1 = 550 N Therefore, if there is no problem with a shaft diameter of 20 mm and a lead of 20 mm (smallest thread minor diameter of 17.5 mm), then the screw shaft diameter of 30 mm should meet the re-quirements. Thus, the following calculations for the buckling load and the permissible compressive and tensile load of the screw shaft are performed while assuming a screw shaft diameter of 20 mm and a lead of 20 mm.

B15-72

511E

Ball Screw

Buckling Load on the Screw Shaft Factor according to the mounting method 2 =20 (see B15-38 ) Since the mounting method for the section between the nut and the bearing, where buckling is to be considered, is “fi xed-fi xed: “ Distance between two mounting surfaces ℓ a =1100 mm (estimate) Screw-shaft thread minor diameter d 1 =17.5 mm

d14

ℓa2

17.54

11002 P1 = η2• ×104 = 20× ×104 = 15500 N

Permissible Compressive and Tensile Load of the Screw Shaft P 2 = 116 × d 1 2 = 116 × 17.5 2 = 35500 N

Thus, the buckling load and the permissible compressive and the tensile load of the screw shaft are at least equal to the maximum axial load. Therefore, a Ball Screw that meets these requirements can be used without a problem.

Studying the Permissible Rotational Speed Maximum Rotational Speed ● Screw shaft diameter: 20 mm; lead: 20 mm Maximum speed V max =1 m/s Lead Ph= 20 mm

Nmax = = 3000 min–1 Vmax×60×103

Ph ● Screw shaft diameter: 20 mm; lead: 40mm Maximum speed V max =1 m/s Lead Ph= 40 mm

Vmax×60×103

Ph Nmax = = 1500 min–1

● Screw shaft diameter: 30mm; lead: 60mm Maximum speed V max =1 m/s Lead Ph= 60 mm

Nmax = = 1000 min–1 Vmax×60×103

Ph


B15-73

511E

Permissible Rotational Speed Determined by the Dangerous Speed of the Screw Shaft Factor according to the mounting method 2 =15.1 (see B15-40 ) Since the mounting method for the section between the nut and the bearing, where dangerous speed is to be considered, is “fi xed-supported: “ Distance between two mounting surfaces ℓ b =1100 mm (estimate)

● Screw shaft diameter: 20 mm; lead: 20 mm and 40 mm Screw-shaft thread minor diameter d 1 =17.5mm

ℓb2

d1

11002 17.5 N1 = λ2× 107 = 15.1× × 107 = 2180 min–1

● Screw shaft diameter: 30mm; lead: 60mm Screw-shaft thread minor diameter d 1 =26.4mm

ℓb2

d1

11002 26.4 N1 = λ2× 107 = 15.1× × 107 = 3294 min–1

Permissible Rotational Speed Determined by the DN Value ● Screw shaft diameter: 20 mm; lead: 20 mm and 40 mm (large lead Ball Screw) Ball center-to-center diameter D=20.75 mm

D 70000

20.75 70000 N2 = = = 3370 min–1

● Screw shaft diameter: 30 mm; lead: 60 mm (large lead Ball Screw) Ball center-to-center diameter D=31.25 mm

D 70000

31.25 70000 N2 = = = 2240 min–1

Thus, with a Ball Screw having a screw shaft diameter of 20 mm and a lead of 20 mm, the maximum rotational speed exceeds the dangerous speed. In contrast, a combination of a screw shaft diameter of 20 mm and a lead of 40 mm, and another of a screw shaft diameter of 30 mm and a lead of 60 mm, meet the dangerous speed and the DN value. Accordingly, a Ball Screw with a screw shaft diameter of 20 mm and a lead of 40 mm, or with a screw shaft diameter of 30 mm and a lead of 60 mm, is selected.

[Selecting a Nut] Selecting a Nut Model Number

Rolled Ball Screw models with a screw shaft diameter of 20 mm and a lead of 40 mm, or with a screw shaft diameter of 30 mm and a lead of 60 mm, are large lead Rolled Ball Screw model WTF variations.

WTF2040-2 (Ca=5.4 kN, C 0 a=13.6 kN) WTF2040-3 (Ca=6.6 kN, C 0 a=17.2 kN) WTF3060-2 (Ca=11.8 kN, C 0 a=30.6 kN) WTF3060-3 (Ca=14.5 kN, C 0 a=38.9 kN)

B15-74

511E

Ball Screw

Studying the Permissible Axial Load Study the permissible axial load of model WTF2040-2 (C 0 a = 13.6 kN). Assuming that this model is used in high-speed transfer equipment and an impact load is applied during deceleration, set the static safety factor (f S ) at 2.5 (see Table1 on B15-47 ).

C0a fS

13.6 2.5 = = 5.44 kN = 5440 N

The obtained permissible axial load is greater than the maximum axial load of 550 N, and therefore, there will be no problem with this model.

Calculating the Travel Distance Maximum speed V max =1 m/s Acceleration time t 1 = 0.15s Deceleration time t 3 = 0.15s

● Travel distance during acceleration

2 2 1×0.15

ℓ1, 4 = ×103 = ×103 = 75 mmVmax• t1

● Travel distance during uniform motion

1×0.15 + 1×0.1522

ℓ2, 5 = ℓS – ×103 = 1000 – ×103 = 850 mmVmax• t1 + Vmax• t3

● Travel distance during deceleration

2 21×0.15

ℓ3, 6 = ×103 = ×103 = 75 mmVmax• t3

Based on the conditions above, the relationship between the applied axial load and the travel dis-tance is shown in the table below.

Motion Applied axial load Fa N (N)

Travel distance ℓ N (mm)

No.1: Duringforward acceleration 550 75

No.2: Duringforward uniform motion 17 850

No.3: Duringforward deceleration –516 75

No.4: Duringbackward acceleration –550 75

No.5: Duringuniform backward motion –17 850

No.6: Duringbackward deceleration 516 75

＊ The subscript (N) indicates a motion number.

Since the load direction (as expressed in positive or negative sign) is reversed with Fa 3 , Fa 4 and Fa 5 , calculate the average axial load in the two directions.


B15-75

511E

Average Axial Load ● Average axial load in the positive direction

Since the load direction varies, calculate the average axial load while assuming Fa 3, 4, 5 = 0N.

Fm1 = = 225 N

3

Fa13× ℓ1 + Fa2

3× ℓ2 + Fa63× ℓ6

ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 + ℓ6

● Average axial load in the negative direction Since the load direction varies, calculate the average axial load while assuming Fa 1, 2, 6 = 0N.

Fm2 = = 225 N

3

Fa3 3× ℓ3 + Fa4

3× ℓ4 + Fa5 3× ℓ5

ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 + ℓ6

Since F m1 = F m2 , assume the average axial load to be F m = F m1 = F m2 = 225 N.

Nominal Life Load factor f W = 1.5 (see Table2 on B15-48 ) Average load F m = 225 N Nominal life L (rev)

L = ×106 Ca ( )

3

fw · Fm

Assumed modelnumber

Dynamic load rating Ca(N)

Nominal life L(rev)

WTF 2040-2 5400 4.1×10 9

WTF 2040-3 6600 7.47×10 9

WTF 3060-2 11800 4.27×10 10

WTF 3060-3 14500 7.93×10 10

B15-76

511E

Ball Screw

Average Revolutions per Minute Number of reciprocations per minute n =8min -1 Stroke ℓ S =1000 mm

● Lead: Ph = 40 mm

Nm = = = 400 min–1

Ph 2×n×ℓs

40 2×8×1000

● Lead: Ph = 60 mm

Ph 2×n×ℓs

60 2×8×1000 Nm = = = 267 min–1

Calculating the Service Life Time on the Basis of the Nominal Life ● WTF2040-2

Nominal life L=4.1×10 9 rev Average revolutions per minute Nm = 400 min -1

60×Nm

L 60×400 4.1×109

Lh = = = 171000 h

● WTF2040-3 Nominal life L=7.47×10 9 rev Average revolutions per minute Nm = 400 min -1

60×Nm

L 60×400 7.47×109

Lh = = = 311000 h


60×Nm

L 60×267

4.27×1010

Lh = = = 2670000 h


60×Nm

L 60×267

7.93×1010

Lh = = = 4950000 h


B15-77

511E

Calculating the Service Life in Travel Distance on the Basis of the Nominal Life ● WTF2040-2

Nominal life L=4.1×10 9 rev Lead Ph= 40 mm L S = L × Ph× 10 -6 = 164000 km

● WTF2040-3 Nominal life L=7.47×10 9 rev Lead Ph= 40 mm L S = L × Ph× 10 -6 = 298800 km



With all the conditions stated above, the following models satisfying the desired service life time of 30,000 hours are selected.

WTF 2040-2 WTF 2040-3 WTF 3060-2 WTF 3060-3

B15-78

511E

Ball Screw

[Studying the Rigidity] Since the conditions for selection do not include rigidity and this element is not particularly neces-sary, it is not described here.

[Studying the Positioning Accuracy] Studying the Lead Angle Accuracy

Accuracy grade C7 was selected in Section [Selecting Lead Angle Accuracy and Axial Clearance] on B15-70 .

C7 (travel distance error: 0.05mm/300mm)

Studying the Axial Clearance Since positioning is performed in a given direction only, axial clearance is not included in the positioning accuracy. As a result, there is no need to study the axial clearance.

WTF2040: axial clearance: 0.1 mm WTF3060: axial clearance: 0.14 mm

Studying the Axial Rigidity Since the load direction does not change, it is unnecessary to study the positioning accuracy on the basis of the axial rigidity.

Studying the Thermal Displacement through Heat Generation Assume the temperature rise during operation to be 5℃. The positioning accuracy based on the temperature rise is obtained as follows:

ℓ = ×t ×ℓ = 12 × 10 –6 × 5 × 1000 = 0.06 mm

Studying the Orientation Change during Traveling Since the ball screw center is 150 mm away from the point where the highest accuracy is required, it is necessary to study the orientation change during traveling. Assume that pitching can be done within 10 seconds because of the structure. The positioning er-ror due to the pitching is obtained as follows:

a = ℓ× sin = 150 × sin (10´´) = 0.007 mm

Thus, the positioning accuracy (p) is obtained as follows:

300 Δp = ± 0.007 + 0.06 = 0.234 mm ±0.05×1000

Since models WTF2040-2, WTF2040-3, WTF3060-2 and WTF3060-3 meet the selection require-ments throughout the studying process in Section [Selecting Lead Angle Accuracy and Axial Clear-ance] on B15-70 to Section [Studying the Positioning Accuracy] on B15-79 , the most compact model WTF2040-2 is selected.


B15-79

511E

[Studying the Rotational Torque] Friction Torque Due to an External Load

The friction toruque is obtained as follows:

2×π×0.9

17×40 2π•η

Fa•Ph T1 = •A = × 1 = 120 N•mm

Torque Due to a Preload on the Ball Screw The Ball Screw is not provided with a preload.

Torque Required for Acceleration Inertial Moment Since the inertial moment per unit length of the screw shaft is 1.23 × 10 -3 kg•cm 2 /mm (see the spec-ifi cation table), the inertial moment of the screw shaft with an overall length of 1200 mm is obtained as follows. J s = 1.23 × 10 –3 × 1200 = 1.48 kg • cm 2 = 1.48 × 10 ‒4 kg • m 2

( )2

40 2×π( )

2 Ph 2×π

J = (m1+m2) •A2×10–6+Js•A2 = (60+20) ×12×10–6+1.48×10–4×12

= 3.39×10–3 kg•m2

Angular acceleration:

60×0.15 2π×1500ω′ = = = 1050 rad/s2

60• t1

2π•Nm

Based on the above, the torque required for acceleration is obtained as follows. T 2 = (J + J m ) ×´ = (3.39 × 10 –3 + 1 × 10 –3 ) × 1050 = 4.61N • m = 4.61 × 10 3 N • mm Therefore, the required torque is specifi ed as follows. During acceleration

T k = T 1 + T 2 = 120 + 4.61×10 3 = 4730 N • mm During uniform motion

T t = T 1 = 120 N • mm During deceleration

T g = T 1 – T 2 = 120 – 4.61×10 3 = – 4490 N • mm

B15-80

511E

Ball Screw

[Studying the Driving Motor] Rotational Speed

Since the Ball Screw lead is selected based on the rated rotational speed of the motor, it is unneces-sary to study the rotational speed of the motor.

Maximum working rotational speed : 1500 min –1 Rated rotational speed of the motor: 3000 min –1

Minimum Feed Amount As with the rotational speed, the Ball Screw lead is selected based on the encoder normally used for an AC servomotor. Therefore, it is unnecessary to study this factor.

Encoder resolution: 1000 p/rev. Doubled: 2000 p/rev

Motor Torque The torque during acceleration calculated in Section [Studying the Rotational Torque] on B15-80 is the required maximum torque.

T max = 4730 N • mm Therefore, the instantaneous maximum torque of the AC servomotor needs to be at least 4,730 N-mm.

Effective Torque Value The selection requirements and the torque calculated in Section [Studying the Rotational Torque] on B15-80 can be expressed as follows. During acceleration:

T k = 4730 N • mm t 1 = 0.15 s

During uniform motion: T t = 120 N • mm t 2 = 0.85 s

During deceleration: T g = 4490 N • mm t 3 = 0.15 s

When stationary: T S = 0 t 4 = 2.6 s

The effective torque is obtained as follows, and the rated torque of the motor must be 1305 N•mm or greater.

2 2 2 2 t4 t3 t2 44902 0 0.15 0.85 2.6 0.85 0.15 0.15

0.15 1202 47302 Ts Tg Tt Tk t1

t4 t3 t2 t1 Trms

1305 N mm


B15-81

511E

Inertial Moment The inertial moment applied to the motor equals to the inertial moment calculated in Section [Studying the Rotational Torque] on B15-80 .

J = 3.39 × 10 –3 kg • m 2 Normally, the motor needs to have an inertial moment at least one tenth of the inertial moment ap-plied to the motor, although the specifi c value varies depending on the motor manufacturer. Therefore, the inertial moment of the AC servomotor must be 3.39 × 10 –4 kg-m 2 or greater. The selection has been completed.

B15-82

511E

Ball Screw

Vertical Conveyance System

[Selection Conditions] Table Mass m 1 =40kg Work Mass m 2 =10kg Stroke length ℓ s = 600mm Maximum speed V max =0.3m/s Acceleration time t 1 = 0.2s Deceleration time t 3 = 0.2s Number of reciprocations per minute n =5min -1 Backlash 0.1mm Positioning accuracy 0.7mm/600mm Positioning accuracy repeatability 0.05mm Minimum feed amount s = 0.01mm/pulse Service life time 20000h Driving motor AC servo motor Rated rotational speed: 3,000 min -1 Inertial moment of the motor J m =5×10 –5 kg•m 2 Reduction gear None (direct coupling) Frictional coeffi cient of the guide surface =0.003 (rolling) Guide surface resistance f=20 N (without load)

600

m1

m2

[Selection Items] Screw shaft diameter Lead Nut model No. Accuracy Axial clearance Screw shaft support method Driving motor


B15-83

511E

[Selecting Lead Angle Accuracy and Axial Clearance] Selecting the Lead Angle Accuracy

To achieve positioning accuracy of 0.7mm/600mm:

600 300

= ±0.7 ±0.35

The lead angle accuracy must be 0.35mm/300 mm or higher. Therefore, the accuracy grade of the Ball Screw (see Table1 on B15-20 ) needs to be C10 (travel distance error: 0.21 mm/300 mm). Accuracy grade C10 is available for low priced, Rolled Ball Screws. Assume that a Rolled Ball Screw is selected.

Selecting the Axial Clearance The required backlashes is 0.1 mm or less. However, since an axial load is constantly applied in a single direction with vertical mount, the axial load does not serve as a backlash no matter how large it is. Therefore, a low price, rolled Ball Screw is selected since there will not be a problem in axial clear-ance.

[Selecting a Screw Shaft] Assuming the Screw Shaft Length

Assume the overall nut length to be 100 mm and the screw shaft end length to be 100 mm. Therefore, the overall length is determined as follows based on the stroke length of 600mm. 600 + 200 = 800 mm Thus, the screw shaft length is assumed to be 800 mm.

Selecting the Lead With the driving motor’s rated rotational speed being 3,000 min –1 and the maximum speed 0.3 m/s, the Ball Screw lead is obtained as follows:

0.3×60×1000 3000

= 6 mm

Therefore, it is necessary to select a type with a lead of 6mm or longer. In addition, the Ball Screw and the motor can be mounted in direct coupling without using a reduc-tion gear. The minimum resolution per revolution of an AC servomotor is obtained based on the resolution of the encoder (1,000 p/rev; 1,500 p/rev) provided as a standard accessory for the AC servomotor, as indicated below. 1000 p/rev(without multiplication) 1500 p/rev(without multiplication) 2000 p/rev(doubled) 3000 p/rev(doubled) 4000 p/rev(quadrupled) 6000 p/rev(quadrupled)

B15-84

511E

Ball Screw

To meet the minimum feed amount of 0.010mm/pulse, which is the selection requirement, the follow-ing should apply. Lead 6mm —— 3000 p/rev 8mm —— 4000 p/rev 10mm —— 1000 p/rev 20mm —— 2000 p/rev 40mm —— 2000 p/rev However, with the lead being 6 mm or 8 mm, the feed distance is 0.002 mm/pulse, and the starting pulse of the controller that issues commands to the motor driver needs to be at least 150 kpps, and the cost of the controller may be higher. In addition, if the lead of the Ball Screw is greater, the torque required for the motor is also greater, and thus the cost will be higher. Therefore, select 10 mm for the Ball Screw lead.

Selecting the Screw Shaft Diameter Those Ball Screw models that meet the lead being 10 mm as described in Section [Selecting Lead Angle Accuracy and Axial Clearance] on B15-84 and Section [Selecting a Screw Shaft] on B15-84 (see Table20 on B15-35 ) are as follows. Shaft diameter Lead 15mm —— 10mm 20mm —— 10mm 25mm —— 10mm Accordingly, the combination of a screw shaft diameter of 15 mm and a lead 10 mm is selected.

Selecting the Screw Shaft Support Method Since the assumed Ball Screw has a stroke length of 600 mm and operates at a maximum speed of 0.3 m/s (Ball Screw rotational speed: 1,800 min -1 ), select the fi xed-supported confi guration for the screw shaft support.


B15-85

511E

Studying the Permissible Axial Load Calculating the Maximum Axial Load

Guide surface resistance f=20 N (without load) Table Mass m 1 =40 kg Work Mass m 2 =10 kg Maximum speed V max =0.3 m/s Acceleration time t 1 = 0.2s

Accordingly, the required values are obtained as follows. Acceleration

α = = 1.5 m/s2Vmax

t1

During upward acceleration: Fa 1 = (m 1 + m 2 ) •g + f + (m 1 + m 2 ) • = 585 N

During upward uniform motion: Fa 2 = (m 1 + m 2 ) •g + f = 510 N

During upward deceleration: Fa 3 = (m 1 + m 2 ) •g + f – (m 1 + m 2 ) • = 435 N

During downward acceleration: Fa 4 = (m 1 + m 2 ) •g – f – (m 1 + m 2 ) • = 395 N

During downward uniform motion: Fa 5 = (m 1 + m 2 ) •g – f = 470 N

During downward deceleration: Fa 6 = (m 1 + m 2 ) •g – f + (m 1 + m 2 ) • = 545 N

Thus, the maximum axial load applied on the Ball Screw is as follows: Fa max = Fa 1 = 585 N

Buckling Load of the Screw Shaft Factor according to the mounting method 2 =20 (see B15-38 ) Since the mounting method for the section between the nut and the bearing, where buckling is to be considered, is “fi xed-fi xed: ” Distance between two mounting surfaces ℓ a =700 mm (estimate) Screw-shaft thread minor diameter d 1 =12.5 mm

d14

ℓa2

12.54

7002 P1 = η2• ×104 = 20× ×104 = 9960 N

Permissible Compressive and Tensile Load of the Screw Shaft P 2 = 116d 1 2 = 116 × 12.5 2 = 18100 N

Thus, the buckling load and the permissible compressive and tensile load of the screw shaft are at least equal to the maximum axial load. Therefore, a Ball Screw that meets these requirements can be used without a problem.

B15-86

511E

Ball Screw

Studying the Permissible Rotational Speed Maximum Rotational Speed ● Screw shaft diameter: 15mm; lead: 10mm

Maximum speed V max =0.3 m/s Lead Ph= 10 mm

Vmax×60×103

Ph Nmax = = 1800 min–1

Permissible Rotational Speed Determined by the Dangerous Speed of the Screw Shaft Factor according to the mounting method 2 =15.1 (see B15-40 ) Since the mounting method for the section between the nut and the bearing, where dangerous speed is to be considered, is “fi xed-supported: ” Distance between two mounting surfaces ℓ b =700 mm (estimate)

● Screw shaft diameter: 15mm; lead: 10mm Screw-shaft thread minor diameter d 1 =12.5 mm

ℓb2

d1

700212.5N1 = λ2× 107 = 15.1× × 107 = 3852 min–1

Permissible Rotational Speed Determined by the DN Value ● Screw shaft diameter: 15mm; lead: 10mm (large lead Ball Screw)

Ball center-to-center diameter D=15.75 mm

D

70000 15.75 70000 N2 = = = 4444 min–1

Thus, the dangerous speed and the DN value of the screw shaft are met.


B15-87

511E

[Selecting a Nut] Selecting a Nut Model Number

The Rolled Ball Screw with a screw shaft diameter of 15 mm and a lead of 10 mm is the following large-lead Rolled Ball Screw model.

BLK1510-5.6 (Ca=9.8 kN, C 0 a=25.2 kN)

Studying the Permissible Axial Load Assuming that an impact load is applied during an acceleration and a deceleration, set the static safety factor (f S ) at 2 (see Table1 on B15-47 ).

C0a fS

25.2 2 Famax = = = 12.6 kN = 12600 N

The obtained permissible axial load is greater than the maximum axial load of 585 N, and therefore, there will be no problem with this model.

Studying the Service Life Calculating the Travel Distance

Maximum speed V max =0.3 m/s Acceleration time t 1 = 0.2s Deceleration time t 3 = 0.2s

● Travel distance during acceleration

2 2 0.3×0.2

ℓ1, 4 = ×103 = ×103 = 30 mmVmax• t1

● Travel distance during uniform motion

0.3×0.2 + 0.3×0.2 2 2

ℓ2, 5 = ℓS – ×103 = 600 – ×103 = 540 mmVmax• t1 + Vmax• t3

● Travel distance during deceleration

2 2 0.3×0.2

ℓ3, 6 = ×103 = ×103 = 30 mmVmax• t3

Based on the conditions above, the relationship between the applied axial load and the travel dis-tance is shown in the table below.

Motion Applied axial load Fa N (N)

Travel distance ℓ N (mm)

No1: During upward acceleration 585 30 No2: During upward uniform motion 510 540

No3: During upward deceleration 435 30 No4: During downward acceleration 395 30

No5: During downward uniform motion 470 540 No6: During downward deceleration 545 30

＊ The subscript (N) indicates a motion number.

B15-88

511E

Ball Screw

Average Axial Load

3

Fm = (Fa13•ℓ1 + Fa2

3•ℓ2 + Fa33•ℓ3 + Fa4

3•ℓ4 + Fa53•ℓ5 + Fa6

3•ℓ6) = 492 N12× ℓS

Nominal Life Dynamic load rating Ca= 9800 N Load factor f W = 1.5 (see Table2 on B15-48 ) Average load F m = 492 N Nominal life L (rev)

L = ×106 = ×106 = 2.34×109 revCa ( )

3

1.5×492 9800 ( )

3

fw · Fm

Average Revolutions per Minute Number of reciprocations per minute n = 5 min -1 Stroke ℓ S =600 mm Lead Ph= 10 mm

Ph 2×n×ℓs

10 2×5×600Nm = = = 600 min–1

Calculating the Service Life Time on the Basis of the Nominal Life Nominal life L=2.34×10 9 rev Average revolutions per minute N m = 600 min -1

L 60×600 2.34×109

Lh = = = 65000 h60•Nm

Calculating the Service Life in Travel Distance on the Basis of the Nominal Life Nominal life L=2.34×10 9 rev Lead Ph= 10 mm L S = L × Ph× 10 -6 = 23400 km

With all the conditions stated above, model BLK1510-5.6 satisfi es the desired service life time of 20,000 hours.


B15-89

511E

[Studying the Rigidity] Since the conditions for selection do not include rigidity and this element is not particularly neces-sary, it is not described here.

[Studying the Positioning Accuracy] Studying the Lead Angle Accuracy

Accuracy grade C10 was selected in Section [Selecting Lead Angle Accuracy and Axial Clearance] on B15-84 .

C10 (travel distance error: 0.21mm/300mm)

Studying the Axial Clearance Since the axial load is constantly present in a given direction only because of vertical mount, there is no need to study the axial clearance.

Studying the Axial Rigidity Since the lead angle accuracy is achieved beyond the required positioning accuracy, there is no need to study the positioning accuracy determined by axial rigidity.

Studying the Thermal Displacement through Heat Generation Since the lead angle accuracy is achieved beyond the required positioning accuracy, there is no need to study the positioning accuracy determined by the heat generation.

Studying the Orientation Change during Traveling Since the lead angle accuracy is achieved at a much higher degree than the required positioning ac-curacy, there is no need to study the positioning accuracy.

[Studying the Rotational Torque] Frictional Torque Due to an External Load

During upward uniform motion:

2×π×η 2×π×0.9

510×10 T1 = = = 900 N•mm Fa2•Ph

During downward uniform motion:

2×π×η 2×π×0.9

470×10 T2 = = = 830 N•mm Fa5•Ph

Torque Due to a Preload on the Ball Screw The Ball Screw is not provided with a preload.

B15-90

511E

Ball Screw

Torque Required for Acceleration Inertial Moment: Since the inertial moment per unit length of the screw shaft is 3.9 × 10 -4 kg•cm 2 /mm (see the speci-fi cation table), the inertial moment of the screw shaft with an overall length of 800mm is obtained as follows. J S = 3.9 × 10 –4 × 800 = 0.31 kg • cm 2 = 0.31 × 10 ‒4 kg • m 2

( )2 10

2×π( )2 Ph

2×πJ = (m1+m2) •A2×10–6+Js•A2 = (40+10) ×12×10–6+0.31×10–4×12

= 1.58×10–4 kg•m2

Angular acceleration:

60• t 60×0.2

2π×1800ω′ = = = 942 rad/s22π•Nmax

Based on the above, the torque required for acceleration is obtained as follows. T 3 = (J + J m ) •´ = (1.58 × 10 –4 + 5 × 10 –5 ) × 942 = 0.2 N•m = 200 N•mm

Therefore, the required torque is specifi ed as follows. During upward acceleration:

T k1 = T 1 + T 3 = 900 + 200 = 1100 N•mm During upward uniform motion:

T t1 = T 1 = 900 N•mm During upward deceleration:

T g1 = T 1 – T 3 = 900 – 200 = 700 N•mm During downward acceleration:

T k2 = 630 N•mm During downward uniform motion:

T t2 = 830 N•mm During downward deceleration:

T g2 = 1030 N•mm


B15-91

511E

[Studying the Driving Motor] Rotational Speed

Since the Ball Screw lead is selected based on the rated rotational speed of the motor, it is unneces-sary to study the rotational speed of the motor.

Maximum working rotational speed : 1800 min –1 Rated rotational speed of the motor: 3000 min –1

Minimum Feed Amount As with the rotational speed, the Ball Screw lead is selected based on the encoder normally used for an AC servomotor. Therefore, it is unnecessary to study this factor.

Encoder resolution: 1000 p/rev.

Motor Torque The torque during acceleration calculated in Section [Studying the Rotational Torque] on B15-90 is the required maximum torque.

T max = T k1 = 1100 N•mm Therefore, the maximum peak torque of the AC servomotor needs to be at least 1100 N-mm.

Effective Torque Value The selection requirements and the torque calculated in Section [Studying the Rotational Torque] on B15-90 can be expressed as follows. During upward acceleration:

T k1 = 1100 N•mm t 1 = 0.2 s

During upward uniform motion: T t1 = 900 N•mm t 2 = 1.8 s

During upward deceleration: T g 1 = 700 N•mm t 3 = 0.2 s

During downward acceleration: T k2 = 630 N•mm t 1 = 0.2 s

During downward uniform motion: T t2 = 830 N•mm t 2 = 1.8 s

During downward deceleration: T g2 = 1030 N•mm t 3 = 0.2 s

When stationary(m 2 =0): T S = 658 N•mm t 4 = 7.6 s

B15-92

511E

Ball Screw

The effective torque is obtained as follows, and the rated torque of the motor must be 743 N•mm or greater.

Trms =

=

Tk12• t1＋Tt1

2• t2＋Tg12• t3＋Tk2

2• t1＋Tt22• t2＋Tg2

2• t3＋Ts2• t4

t1＋ t2＋ t3＋ t1＋ t2＋ t3＋ t4

11002×0.2＋9002×1.8＋7002×0.2＋6302×0.2＋8302×1.8＋10302×0.2＋6582×7.6 0.2＋1.8＋0.2＋0.2＋1.8＋0.2＋7.6

= 743 N•mm

Inertial Moment The inertial moment applied to the motor equals to the inertial moment calculated in Section [Studying the Rotational Torque] on B15-90 .

J = 1.58 × 10 –4 kg • m 2 Normally, the motor needs to have an inertial moment at least one tenth of the inertial moment ap-plied to the motor, although the specifi c value varies depending on the motor manufacturer. Therefore, the inertial moment of the AC servomotor must be 1.58 × 10 –5 kg-m 2 or greater. The selection has been completed.


B15-93

511E

B15-94

511E

examples of selecting a ball screw - thk · pdf filescrew shaft diameter of 30 mm and a lead...

Documents